Answer:
The first 50 elements along with their valences are given below :
1. Hydrogen = 1
2. Helium = 0
3. Lithium = 1
4. Beryllium = 2
5. Boron = 3
6. Carbon = 4
7. Nitrogen = 3
8. Oxygen = 2
9. Fluorine = 1
10. Neon = 0
11. Sodium = 1
12. Magnesium = 2
13. Aluminium = 3
14. Silicon = 4
15. Phosphorus = 3
16. Sulphur = 2
17. Chlorine = 1
18. Argon = 0
19. Potassium = 1
20. Calcium = 2
21. Scandiun = 3
22. Titanium = 3
23. Vanadium = 4
24. Chromium = 3
25. Manganese = 4
26. Iron = 2
27. Cobalt = 2
28. Nickel = 2
29. Copper = 2
30. Zinc = 2
31. Gallium = 3
32. Germanium = 4
33. Arsenic = 3
34. Selenium = 2
35. Bromine = 1
36. Krypton = 0
37. Rubidium = 1
38. Strontium = 2
39. Yttrium = 3
40. Zirconium = 4
41. Niobium = 3
42. Molybdenum = 3
43. Technetium = 7
44. Ruthenium = 4
45. Rhodium = 3
46. Palladium = 4
47. Sliver = 1
48. Cadmium = 2
49. Indium = 3
50. Tin = 4
<u>Note</u> :
An element like Iron, copper can have more than one valencies.
The answer u are looking for is b
The molar mass of the compound potassium nitrate, KNO3 is equal to 101.1032 g/mol. Then, we determine the number of moles present in the given amount,
n = 11.75g / (101.1032 g/mol) = 0.116 mol
Then, molarity is calculated by dividing the number of moles by the volume of the solution. The answer is therefore 0.058 M.
Answer:
665 g
Explanation:
Let's consider the following thermochemical equation.
2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(l), ΔH°rxn= –5,314 kJ/mol
According to this equation, 5,314 kJ are released per 8 moles of CO₂. The moles produced when 1.00 × 10⁴ kJ are released are:
-1.00 × 10⁴ kJ × (8 mol CO₂/-5,314 kJ) = 15.1 mol CO₂
The molar mass of CO₂ is 44.01 g/mol. The mass corresponding to 15.1 moles is:
15.1 mol × 44.01 g/mol = 665 g
This is because oxygen (2.8.6) requires two electrons on its valence shell to attain stable configuration (2.8.8). Hydrogen (1) on the other hand requires one electron on its valence shell to attain stable configuration (2). Therefore in a covalent bond, it requires two hydrogen and one oxygen to share electrons and achieve stable configuration.
